Making a Box

[SydneyRoverP6B]

You are given the task to make a cardboard box without a lid. The volume of the box is to be 32,000 cubic cm. The only constraint is that that you are to use the least amount of carboard possible with each of the five faces completely made from the cardboard without any holes or pieces missing. No overlapping is required. What will be the dimensions of the box?

Ron.

 

[Pilkie]

Whatever shape the box,be it cube,rectangle,oblong etc the volume will stay the same!
So it could be 32000cm long x 1cm high x 1cm wide!
or
l=32cm x w=100cm x h=10cm!
And it would also depend on the size of the piece of cardboard given to you for the task! :wink:

 

[quattro]

With just a smidge of trial and error, I have 40w x 40l x 20h

Working on the basis of a cube being the shape with the largest volume for its area, when working with 90 degree six sided objects, I have assumed that the above measurements, (half a cube) would be the best formulation.

If you were counting the top as well, then it would be simply root 3 of 32,000

Richard

Area would be 4,800cm2

 

[darth sidious]

With just a smidge of trial and error, I have 40w x 40l x 20h

Working on the basis of a cube being the shape with the largest volume for its area, when working with 90 degree six sided objects, I have assumed that the above measurements, (half a cube) would be the best formulation.

If you were counting the top as well, then it would be simply root 3 of 32,000

Richard

Area would be 4,800cm2

I have to concur, after some trial and error with excel, Richard's solution seems the optimum in terms of economy of cardboard!

 

[SydneyRoverP6B]

pilkie wrote,..

Whatever shape the box,be it cube,rectangle,oblong etc the volume will stay the same!
So it could be 32000cm long x 1cm high x 1cm wide!
or
l=32cm x w=100cm x h=10cm!
And it would also depend on the size of the piece of cardboard given to you for the task!

Hello Pilkie,

I am glad that you are enjoying the problems... On your first point regarding the shape and the resultant volume..that is correct as the required volume is to be 32,000 cubic cm as stated in the question. On your second point,..the question requires that the least amount of cardboard be used so that the surface area is as small as possible. If a person is paying for the cardboard to be used, then the less you can use for the required volume, the more money you will save.... :wink:

quattro wrote,...

With just a smidge of trial and error, I have 40w x 40l x 20h

Darth Sidious wrote....

I have to concur, after some trial and error with excel, Richard's solution seems the optimum in terms of economy of cardboard!

Hello Richard and Darth,

Yes indeed, that is the correct answer that will ensure that the least amount of cardboard is used while meeting the requirement of the required volume.

As an alternative to the trial and error method you can derive an expression for the surface area of the prism and then use partial derivatives to acertain the required lengths.

Well done to all....

Ron.

 

[quattro]

As an alternative to the trial and error method you can derive an expression for the surface area of the prism and then use partial derivatives to acertain the required lengths.

No I couldn't :?: